(D) From the ideal gas equation,$PV = nRT$,where $n$ is the number of moles.Since $V$,$R$,and $T$ are constant for both vessels,we have $P \propto n$.

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(D) From the ideal gas equation,$PV = nRT$,where $n$ is the number of moles.Since $V$,$R$,and $T$ are constant for both vessels,we have $P \propto n$.Given the ratio of pressures is $\frac{P_1}{P_2} = \frac{1}{2}$,it follows that the ratio of the number of moles (and thus the number of molecules) is $\frac{n_1}{n_2} = \frac{1}{2}$.Also,the root mean square velocity is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.Since $T$ and $M$ are the same for both vessels,the ratio of $v_{rms}$ is $1 : 1$.Therefore,both statements $(B)$ and $(C)$ are correct.

Class 11 Physics Kinetic Theory of Gases Mix Examples-Kinetic Theory of Gases

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Two vessels of the same volume contain the same gas at the same temperature. If the pressure in the vessels is in the ratio of $1 : 2$,then

A
the ratio of the average kinetic energy is $1 : 2$
B
the ratio of the root mean square velocity is $1 : 1$
C
the ratio of number of molecules is $1 : 2$
D
Both $(B)$ and $(C)$

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Mix Examples-Kinetic Theory of Gases

An ideal gas of molar mass $M$ is contained in a very tall vertical cylindrical column in a uniform gravitational field. Assuming the gas temperature to be $T$, the height at which the centre of gravity of the gas is located is (where $R$ is the universal gas constant).

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At the top of a mountain,a thermometer reads $7^{\circ}C$ and a barometer reads $70 \ cm-Hg$. At the base of the mountain,the thermometer reads $27^{\circ}C$ and the barometer reads $76 \ cm-Hg$. The ratio of the density of air at the top of the mountain to the density of air at the base is:

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For a diatomic gas,the change in internal energy at constant pressure and the change in internal energy for a unit temperature change are $U_1$ and $U_2$ respectively. Then $U_1 : U_2$ is equal to:

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$125 \, ml$ of gas $A$ at $0.60 \, atm$ and $150 \, ml$ of gas $B$ at $0.80 \, atm$ pressure at the same temperature are filled in a vessel of $1 \, L$ volume. What will be the total pressure of the mixture at the same temperature in $atm$?

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An ideal gas $(\gamma = 1.5)$ is expanded adiabatically. How many times must the gas be expanded to reduce the root mean square velocity of the molecules $2.0$ times?

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Two vessels of the same volume contain the same gas at the same temperature. If the pressure in the vessels is in the ratio of $1 : 2$,then

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An ideal gas of molar mass $M$ is contained in a very tall vertical cylindrical column in a uniform gravitational field. Assuming the gas temperature to be $T$, the height at which the centre of gravity of the gas is located is (where $R$ is the universal gas constant).

At the top of a mountain,a thermometer reads $7^{\circ}C$ and a barometer reads $70 \ cm-Hg$. At the base of the mountain,the thermometer reads $27^{\circ}C$ and the barometer reads $76 \ cm-Hg$. The ratio of the density of air at the top of the mountain to the density of air at the base is:

For a diatomic gas,the change in internal energy at constant pressure and the change in internal energy for a unit temperature change are $U_1$ and $U_2$ respectively. Then $U_1 : U_2$ is equal to:

$125 \, ml$ of gas $A$ at $0.60 \, atm$ and $150 \, ml$ of gas $B$ at $0.80 \, atm$ pressure at the same temperature are filled in a vessel of $1 \, L$ volume. What will be the total pressure of the mixture at the same temperature in $atm$?

An ideal gas $(\gamma = 1.5)$ is expanded adiabatically. How many times must the gas be expanded to reduce the root mean square velocity of the molecules $2.0$ times?

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